Can someone post the solution?

Let $a_i$ be the index of the rightmost black cell in row $i$ with $a_i = 0$ if there are no black cells in row $i.$ Then all the cells to its left are black and all the cells to its right are white. We claim that a board achieves$^1$ maximum beauty when $a_i = n-i,$ for which the beauty is $2\sum\limits_{i=1}^n i(n-i) = 2\binom{n+1}{3}.$ When $a_i$ decreases by $1$ and $a_i \ne^* a_{i+1}$ or $i = n,$ the change in contribution from contribution from row $i$ is $(a_i - 1)(n-a_i+1) - a_i(n-a_i) = 2a_i - (n+1).$ However, the contribution from column $a_i$ is $(i-1)(n-i+1) - i(n-i) = 2i-(n+1).$ Since $a_{i-1} \ge a_i > a_{i-1},$ there is no net change in beauty elsewhere, and the total change is $2(a_i+i-(n+1)).$ Thus, we can decrease $a_n, a_{n-1}, \dots, a_1$ in that order (the order is so that $*$ is satisfied) without decreasing beauty until $a_i \le n-i$ for all $i.$ Similarly, if $a_i$ increases by $1$ and $a_i \ne a_{i-1}$ or $i=1,$ the net change in beauty is $(a_i+i)(n-(a_i+1)) - a_i(n-a_i) + i(n-i) - (i-1)(n-i+1) = 2(n-a_i-i),$ so we can increase $a_n, a_{n-1}, \dots, a_1$ in order without decreasing beauty until $a_i \ge n-i$ for all $i.$ This means that from any configuration, we can reach the board where $a_i = n-i$ for all $i$ without decreasing beauty, which means that the maximum occurs$^1$ when $a_i = n-i$ for all $i$ as desired. $^1$ The maximum also occurs when $a_i = n+1-i,$ which is why were careful to say "when" instead of "only when."

Call such a pair $(u,v)$ a "pr0" pair . Let $a_i$ denote the number of black cells in the $i$-th row . Note that since the board is convex so $i \leq j \implies a_i \geq a_j$ So ,$$ \# \text { pr0 pairs with both cells in same row } = \sum_{i=1}^{n} a_i(n-a_i)$$. Next note that number of columns with atleast $i$ black cells is atleast $a_i$ . Hence $\#$ columns with exactly $i$ black cells is $(a_i-a_{i+1})$ So $$ \# \text { pr0 pairs with both cells in same column } = \sum_{i=1}^{n} i(n-i)(a_i-a_{i+1}$$. So we want to maximize $$ \mathcal S = \sum_{i=1}^{n} i(n-i)(a_i-a_{i+1}+a_i(n-a_i) = \sum_{i=1}^{n} a_i(2n+1-2i-a_i)$$ Consider the function $f(a_i)= a_i(2n+1-2i-a_i)$ for fixed $i$ . Note that the maximum value of $f$ is attained at $a_i =\{n-i,n+1-i\}$ .It is easy to check $a_i \leq a_{i+1}$ so the board is convex . So the desired sum is := $$\sum_{i=1}^n (n-i)(n-i+1) = \sum_{i=1}^n i^2-i= \boxed{\frac {n^3-n}{6}}$$

Interesting Problem. Let $r_i$ be the number of black cells in row $i$. Similarly, let $c_i$ be the number of black cells in column $i$. The convex condition of board is equivalent to: $r_1 \ge r_2 \ge \dots \ge r_n$ and $c_1 \ge c_2 \ge c_3 \dots \ge c_n$. (From the fact that for every black colored cell, both the cell directly to the left of it and the cell directly above it are also colored black.) $c_j = \sum_{r_i \ge j} 1$. (Since the $j$th column counts the number of rows having at least $j$ black cells.) This means that $c_i$, $1 \le i \le n$ is uniquely determined by $r_1, r_2, \dots, r_n$. Now, the beauty of the board having $\{ r_1, r_2, \dots, r_n \}$ number of black squares is \[ \sum_{i = 1}^n r_i (n - r_i) + \sum_{j = 1}^n c_j (n - c_j) \]For the sake of convenience, define $r_0 = c_0 = n$ and $r_{n + 1} = c_{n + 1} = 0$. Claim. The number of $i$ in $\{ c_1, c_2, \dots, c_n \}$, where $0 \le i \le n$ is $r_i - r_{i + 1}$. Proof. Well known. Thus, we need to maximize \[ \sum_{i = 1}^n r_i (n - r_i) + \sum_{j = 1}^n (r_j - r_{j + 1})j(n - j) \]We have \begin{align*} & \ \sum_{i = 0}^n r_i (n - r_i) + \sum_{j = 0}^n (r_j - r_{j + 1} )j(n - j) \\ &= n \sum_{i = 0}^n r_i - \sum_{i = 0}^n r_i^2 + n \sum_{j = 0}^n j(r_j - r_{j + 1}) - \sum_{j = 0}^n j^2 (r_j - r_{j + 1}) \\ &= n \sum_{i = 0}^n r_i - \sum_{i = 0}^n r_i^2 + n \sum_{j = 1}^n (j + 1 - j) r_j - \sum_{j = 0}^{n - 1} ((j + 1)^2 - j^2) r_{j + 1} \\ &= n \sum_{i = 0}^n r_i - \sum_{i = 0}^n r_i^2 + n \sum_{i = 1}^n r_i - \sum_{j = 1}^{n} (2j - 1) r_j \\ &= 2n \sum_{i = 1}^n r_i - \sum_{i = 1}^n r_i^2 - \sum_{i = 1}^n (2i - 1)r_i \\ &=\sum_{i = 1}^n r_i(2n - r_i - 2i + 1) \end{align*}Thus, we want to maximize $\sum_{i = 1}^n r_i (2n - r_i - 2i + 1)$ for all possible $r_1 \ge r_2 \ge \dots \ge r_n$. Define $f(r_i) = r_i (2n - 2i + 1 - r_i)$. Now, we need to maximize $f(r_i)$ for each $i$ by choosing the right value of $r_i$. Now, $f(r_i)$ is maximized when $r_i = \{ n - i, n + 1 - i \}$. Since $n + 1 - i \ge r_i \ge n - i \ge r_{i - 1}$. So, these choices are valid. Thus, the maximum value is attained when $r_i \in \{ n - i, n + 1 - i \}$ for every $1 \le i \le n$, which gives \[ \sum_{i = 1}^n (n - i)(n - i + 1) = \sum_{i = 0}^{n - 1} i(i + 1) = \frac{n^3 - n}{6} \]